+0  
 
0
32
2
avatar

Suppose the roots of the polynomial \(x^2 - mx + n\) are positive prime integers (not necessarily distinct). Given that \(m < 20\) , how many possible values of \(m\) are there? (woops sorry I forgot the m)

 Mar 1, 2019
edited by Guest  Mar 1, 2019
 #1
avatar+17331 
+2

I THINK you meant how many possible values of 'n' are there?

 

Since  the two roots are positive....

this means the solution looks like this:

 

(x-a)(x-b)        where  a and  b   are the roots   

 

a and b must ADD to m   and must multiply to n   AND m<20      or   <=19   and POSITIVE integers AND PRIME

 

 

primes below   19

3  5   7   11  13   17   19     Two of these added together  must be <= to 19

 

3 +  3,7,11,13

5 + 5,7,11,13

7+ 7,11

I think that is all the combos   ====>   TEN   possible values of 'n'   (only 6 values of 'm')

 Mar 1, 2019
edited by ElectricPavlov  Mar 1, 2019
 #2
avatar
0

Thank you so much, sorry I did mean n, thank you!!!

Guest Mar 1, 2019

16 Online Users

avatar
avatar